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t <br />A5; A <br />�6��_CUMVE_ TABLES. <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE -CURVE TABLES <br />Table I. contains Tangents and Externals to a 1' curve. Tan. and <br />Ext. to any other radius may be found nearly enough, by dividing the Tan. <br />,or Ext. opposite the giveriCentral Angle by the given degree of curve <br />To find Deg. of Curve, having the Central Angle and Tangent: <br />Divide Tan. opposite the given Central Angle by the given Tangent. <br />, To find Deg. of Curve, having the Central Angle and External: <br />Divide Ext' opposite the given Central Angle by the given External. <br />To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. <br />or Ext. of twice the given -angle divided by the radius of a I' curve will <br />be the Nat. Tan. or Nat. Ex.:Sec. <br />EXAMPLE <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or I. P. =23' 20' to the R. at Station <br />542+72. <br />Ext. in Tab. I opposite 23" 20' = 120.87 <br />120.87 —'.12=10.07., Say.a"10° Curve. <br />Tan. in Tab. I opp. 236 20'=1183.1 <br />1183.1=10=118.31: <br />Correction for* A. 23" 20' for a 10° Cur. =0.16 <br />118.31 +0.16 —_ 118.47 = corrected Tangent. <br />(If corrected Ekt. is required find in same way) <br />Ang. 23* 20'=23.33°=10=2.3333=L. C. <br />2" 192" = def. for sta. 542 1. P. = sta.. 542+72 <br />40 492"= " " " - +50 Tan. = 1 .18.47 <br />7" it a a 543 <br />9' 49 +50 B. C. = sta.' 541+53.53 <br />01 L C = 2 .33.33 <br />2 <br />11" 4 543+ <br />86.86 8. C. = Sta. 5431+86.86 <br />100-53.53=46.47X3'(def. for 1 ft. o ' f 10° ' Cur.).= 139AY7 <br />2'49""=def. for sta. 542. <br />Def. for 50 ft. =2" 30' for a 10° Curve. <br />Def. for 36.86 ft. = Ic' 5015' for a 10° Curve. <br />> <br />40� 01 Z' <br />77— <br />05 <br />0 > <br />0 <br />0077 <br />01 <br />6 <br />1Y. <br />A5; A <br />�6��_CUMVE_ TABLES. <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE -CURVE TABLES <br />Table I. contains Tangents and Externals to a 1' curve. Tan. and <br />Ext. to any other radius may be found nearly enough, by dividing the Tan. <br />,or Ext. opposite the giveriCentral Angle by the given degree of curve <br />To find Deg. of Curve, having the Central Angle and Tangent: <br />Divide Tan. opposite the given Central Angle by the given Tangent. <br />, To find Deg. of Curve, having the Central Angle and External: <br />Divide Ext' opposite the given Central Angle by the given External. <br />To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. <br />or Ext. of twice the given -angle divided by the radius of a I' curve will <br />be the Nat. Tan. or Nat. Ex.:Sec. <br />EXAMPLE <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or I. P. =23' 20' to the R. at Station <br />542+72. <br />Ext. in Tab. I opposite 23" 20' = 120.87 <br />120.87 —'.12=10.07., Say.a"10° Curve. <br />Tan. in Tab. I opp. 236 20'=1183.1 <br />1183.1=10=118.31: <br />Correction for* A. 23" 20' for a 10° Cur. =0.16 <br />118.31 +0.16 —_ 118.47 = corrected Tangent. <br />(If corrected Ekt. is required find in same way) <br />Ang. 23* 20'=23.33°=10=2.3333=L. C. <br />2" 192" = def. for sta. 542 1. P. = sta.. 542+72 <br />40 492"= " " " - +50 Tan. = 1 .18.47 <br />7" it a a 543 <br />9' 49 +50 B. C. = sta.' 541+53.53 <br />01 L C = 2 .33.33 <br />2 <br />11" 4 543+ <br />86.86 8. C. = Sta. 5431+86.86 <br />100-53.53=46.47X3'(def. for 1 ft. o ' f 10° ' Cur.).= 139AY7 <br />2'49""=def. for sta. 542. <br />Def. for 50 ft. =2" 30' for a 10° Curve. <br />Def. for 36.86 ft. = Ic' 5015' for a 10° Curve. <br />> <br />